ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.6

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions Ex 1.6

Question 1.
For each binary operation ‘*‘ defined below, determine whether * commutative and whether * is associative :
(i) On Z, defined by a * b = a – b (NCERT)
(ii) On Q, defined by a * b = \(\frac{ab}{2}\) (NCERT)
(iii) On Q, defined by a * b = ab – 1
(iv) On R, defined by a * b = 2a + 2b
(v) On Q defined by a* b = a – b + ab (NCERT Exemplar)
(vi) On R, defined by a * b = b
(vii) On R – {- 1} defined by a * b = \(\frac{a}{b+1}\) (NCERT)
(viii) On N, defined by a * b = 1
(ix) On N , defined by a * b = a³ + b³.
Solution:
(i) Given binary operation * on Q defined by
a * b = a – b ∀ a, b ∈ Q

Commutativity:
∀ a, b ∈ Q
a * b = a – b ≠ b – a = b * a
∴ a * b ≠ b * a∀ a, b ∈ Q
Thus * is not a commutative binary operation.

Associativity: ∀ a, b, c ∈ Q, we have
(a * b) * c = (a – b) * c
= (a – b) – c
= a – b – c
and a * (b * c) = a * (b – c)
= a – (b – c)
= a – b + c
∴ (a + b) * c ≠ a * (b * c) ∀ a, b, c ∈ Q
Thus * is not associative binary operation on Q.

(ii) Operation O on Q defined by aob = \(\frac{ab}{2}\) ∀ a, b ∈ Q

Commutativity : ∀ a, b ∈ Q
aob = \(\frac{ab}{2}\) = \(\frac{ab}{2}\) = boa ∀ a, b ∈ Q
[Since commutativity hold under multiplication ∴ ab = ba]
Thus ‘o’ is commutative on Q.

Associativity : ∀ a, b, c ∈ Q, we have
(aob) oc = \(\frac{ab}{2}\) oc
= \(\frac{\frac{a b}{2} \cdot c}{2}=\frac{a b c}{4}\)

ao (boc) = a_o (\(\frac{ab}{2}\))
= \(\frac{a\left(\frac{b c}{2}\right)}{2}=\frac{a b c}{4}\)

∴ (aob) oc = ao (boc) ∀ a, b ∈ Q
Thus o is associative binary operation on Q.

(iii) Given binary operation * on Q defined by a

* b = ab – 1 ∀ a, b ∈ Q,
a * b = ab – 1
= ba – 1
= b * a [∵ ∀ a, b ∈ Q, ab = ba]
Thus * be commutative on Q.

∀ a, b, c ∈ Q, we have
(a * b) * c = (ab – 1) * c
= (ab – 1) c – 1
= abc – c – 1

and a* (b * c) = a* (bc – 1)
= a (bc – 1) – 1
= abc – a – 1
Clearly (a * b) * c ≠ a * (b * c)
Thus * is not associative on Q.

(iv) Given binary operation * on R defined by
a* b = 2a + 2b
∀ a, b ∈ R, a * b = 2a + 2b
= 2b + 2a
= b * a
[∵ a + b = b + a ∀ a, b ∈ R]
Thus * is commutative on R.
∀ a, b, c ∈ R, we have
(a * b) * c = (2a + 2b) * c
= 2 (2a + 2b) + 2c
= 4a + 4b + 2c
and a * (b * c) = a * (2b + 2c)
= 2a + 2 (26 + 2c)
= 2a + 4b + 4c
Clearly (a * b) * c ≠ a* (b * c)
Thus * is not associative on R.

(v) Given binaiy operation * on Q defined by a * b = a – b + ab
∀ a, b ∈ Q, a * b = a – b + ab
and b * a = b – a + ab
Clearly a * b ≠ b * a
Thus, * is not commutative on Q.
∀ a, b, c ∈ Q,
(a * b) * c = (a – b + ab) * c
= (a – b + ab) – c + (a – b + ab) c
= a – b + ab – c + ac – bc + abc
= a – b – c + ab + ac – bc + abc
a * (b * c) = a* (b – c + bc)
= a – (b – c + bc) + a (b – c + bc)
= a – b + c – bc + ab – ac + abc
Clearly (a * b) * c ≠ a * (b * c)
Thus * is not associative on Q.

(vi) Given binary relations * on R defined by a * b = b
∀ a, b ∈ R, a * b = b and b * a = a
∴ a * b ≠ b * a (for b ≠ a)
Thus * is not commutative on R.
∀ a, b, c ∈ R, (a * b) * c = b * c = c
and a * (b * c) = a * c = c
∴ (a * b) * c = a * (b * c) ∀ a, b, c ∈ R
Thus, * is associative on R.

(vii) Given binary relation * on R – {- 1} defined by
a * b = \(\frac{a}{b+1}\)
∀ a, b ∈ R – {- 1}, a * b = \(\frac{a}{b+1}\)
and b * a = \(\frac{b}{a+1}\)
Clearly a * b ≠ b * a [for a ≠ b]
e.g. Since 1, 2 ∈ R – {- 1}
s.t. 1 * 2 = \(\frac{1}{2+1}=\frac{1}{3}\)
and 2 * 1 = \(\frac{2}{1+1}\) = 1
∴ 1 * 2 ≠ 2 * 1
Thus * is not commutative on R – {- 1}.
∀ a, b, c ∈ R – {- 1}, we have
(a * b) * c = \(\left(\frac{a}{b+1}\right)\) * c
= \(\frac{a}{\frac{b+1}{c+1}}\)
= \(\frac{a}{(b+1)(c+1)}\)
and a * (b * c) = a * \(\left(\frac{b}{c+1}\right)\)
= \(\frac{a}{\frac{b}{c+1}+1}\)
= \(\frac{a(c+1)}{b+c+1}\)
Clearly (a * b) * c ≠ a * (b * c)
e.g. 1, 2, 3 ∈ R – {- 1} * 3
(1 * 2) \(\frac{1}{2+1}\) * 3 = \(\frac{1}{3}\) * 3
= \(\frac{1}{\frac{3}{3+1}}=\frac{1}{12}\)
and 1 * (2 * 3) = 1 * \(\frac{2}{3+1}\) = 1 * \(\frac{1}{2}\)
1 2 3
(1 * 2) * 3 * 1 * (2 * 3)
Thus, * is not associative on R – {- 1}.

(viii) Given binary operation * on N defined by a * b = 1 ∀ a, b ∈ N
Commutativity:
∀ a, b ∈ N, we have
a * b = 1 = b * a
∴ * is a commutative binary operation on N.

Associativity :
∀ a, b, c ∈ N, we have
(a * b) * c = 1 * c = 1
[since 1, c both are natural numbers]
and a * (b * c) = a * 1 = 1
[∵ a, 1 ∈ N ∴ a * 1 = 1]
Thus, (a * b) * c = a * (b * c)
∴ * be a associative binary operation on N.

(ix) Given binary relation on N defined by
a * b = a³ + b³
∀ a, b ∈ N,
a * b = a³ + b³
= b³ + a³ = b * a
[∵ a + b = b + a ∀ a, b ∈ N]
∴ * is commutative on N.
∀ a, b, c ∈ N, (a * b) * c = (a³ + b³ ) * c
= (a³ + b³ )³ + c³
and a * (b * c) = (a³ + b³) * c
= a³ + (b³ + c³)³
Clearly (a * b) * a ≠ 0 * (b * c)
Thus * is not associative on N.

Question 2.
(i) Is the binary operation ‘*’ on the set R, given by a * b = \(\frac{a+b}{2}\) for all a, b ∈ R, commutative ?
(ii) Is the above binary operation ‘*’ associative ?
Solution:
(i) Given binary operation * on Q defined by
a * b = \(\frac{a+b}{2}\) ∀a, b ∈ Q

Commutativit :
∀a, b ∈ Q, we have
a * b = \(\frac{a+b}{2}=\frac{b+a}{2}\) = b * a
[∵ addition is commutative on Q]
∴ * is a commutative binary operation on Q.

(ii) Associativity : ∀a, b, c ∈ Q we have
(a * b) * c = \(\left(\frac{a+b}{2}\right)\) * c
= b * a
[∀a, b ∈ Q then \(\frac{a+b}{2}\) ∈ Q]
= \(\frac{\frac{a+b}{2}+c}{2}\)
= \(\frac{a+b+2 c}{4}\)
and a* (b * c) = a * \(\frac{b+c}{2}\) =
= \(\frac{a+\frac{b+c}{2}}{2}\)
= \(\frac{2 a+b+c}{2}\)
∴ (a * b) * c ≠ a * (b * c) ∀a, b, c ∈ Q
Thus * is not associative on Q.

Question 3.
Let ‘*’ be a binary operation on the set Q of rational numbers given as a * b = (2a – b)², a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3?
Solution:
Given binary operation * on Q defined by
a * b = (2a – b)² ∀a, b, c ∈ Q
3 * 5 = (2 × 3 – 5)²
= (6 – 5)² = 1² = 1
5 * 3 = (2 × 5 – 3)²
= (10 – 3)² = 7² = 49
∴ 3 * 5 ≠ 5 * 3.

Question 4.
Let * be a binary operation on Q as given below :
(i) a * b = a – b
(ii) a * b = (a – b)²
(iii) a* b = a + ab (NCERT Exemplar)
Find which of the above binary operations are commutative and which are associative. (NCERT)
Solution:
(i) Given binary operation * on Q defined by
a * b = a – b ∀a, b, c ∈ Q,
a * b = a – b = – (b – a)
= – (b * a)
i.e. a * b ≠ b * a
Thus, * is not commutative on Q.
∀a, b, c ∈ Q, we have
(a * b) * c = (a – b) * c
= a – b – c
and a * (b * c) = a * (b – c)
= a – (b – c)
= a – b + c
Clearly (a * b) * c ≠ a * (b * c)
∴ * is not associative on Q.

(ii) Operation * on Q defined by a * b = (a – b)² ∀a, b ∈ Q
Commutativity :
∀a, b ∈ Q, we have
a * b = (a – b)² = (b – a)² = b * a
Thus * is commutative on Q.

Associativity :
∀a, b, c ∈ Q, we have
(a * b) * c = (a – b)² * c
= [(a – b)² – c]²
a * (b * c) = a * (b – c)²
= [a – (b – c)²]²
Clearly (a* b) * c ≠ a * (b * c) ∀a, b, c ∈ Q
Thus * is not an associative binary operation on Q.

(iii) Operation * on Q defined by a * b = a + ab ∀a, b ∈ Q

Commutativity:
∀a, b ∈ Q
a * b = a + ab
and b * a = b + ba
a * b ≠ b * a ∀a, b ∈ Q
[e.g. : 2 *3 = 2 + 6 = 8
and 3 * 2 = 3 + 6 = 9
∴ 2 * 3 ≠ 3 * 2]
Thus * is not commutative binary operation on Q.

Associativity : ∀a, b, c ∈ Q
(a * b) * c = (a + ab) * c
= (a + ab) + (a + ab) c
= a + ab + ac + abc …………..(1)
a * (b * c) = a * (b + bc)
= a + a (b + bc)
= a + ab + abc ……………(2)
From (1) and (2) ; we have
a * (b * c) ≠ (a * b) * c ∀a, b, c ∈ Q
Thus * is not associative binary operation on Q.

Question 5.
Let ‘*’ be a binary operation on Q , ab defined by a * b = \(\frac{ab}{4}\).
(i) Show that the operation * is commutative as well as associative.
(ii) It has 4 as identity element.
(iii) Every non-zero element is invertible. If a * 0 e Q, find its inverse.
Solution:
Operation * on Q defined by a * b = \(\frac{ab}{4}\)
∀a, b ∈ Q
Commutativit: ∀a, b ∈ Q, we have
a * b = \(\frac{ab}{4}\) and
b * a = \(\frac{ba}{4}\) = \(\frac{ab}{4}\)
[Since commutativity holds under multiplication ∴ ab = ba]
Thus, a * b = b * a
∴ * is commutative on Q.

Associativity:
∀a, b, c ∈ Q
(a * b) * c = (\(\frac{ab}{4}\)) * c
= \(\frac{\frac{a b}{4} \times c}{4}=\frac{a b c}{16}\)

and a* (b * c) = a * \(\frac{bc}{4}\)
= \(\frac{a \times \frac{b c}{4}}{4}=\frac{a b c}{16}\)
Clearly (a * b) * c = a * (b – c)
Thus * is associative on Q.

(ii) Let e be the identity element in Q w.r.t. binary operation *.
∴ a * e = a = e * a ∀a ∈ Q
⇒ \(\frac{a e}{4}=a=\frac{e a}{4}\) ∀a ∈ Q
⇒ e = 4
Thus, 4 be the identity element of Q.

(iii) Let a ≠ 0 be any arbitrary element of Q and
let x be the inverse of a
∴ a * x = e = x * a
⇒ \(\frac{ax}{4}\)= 4 and \(\frac{xa}{4}\) = 4
⇒ x = \(\frac{16}{a}\), where a ≠ 0
i.e., x = \(\frac{16}{a}\) ∈ Q [∵ a ≠ 0]
Thus every element of Q be invertible and \(\frac{16}{a}\) be the inverse of a ≠ 0 ∈ Q.

Question 6.
Determine whether or not each of the k definition of ‘*’ given below gives a binary operation. In the event that ‘*’ is not a binary operation, give justification.
(i) On Z⁺, defined by a * b = a – b where Z⁺ is the set of non-negative integers.
(ii) On Z⁺, defined by a * b = ab
(iii) On Z⁺, defined by a * b = |a – b|
(iv) On Z⁺, defined by a * b = a
(v) On Z⁺, defined by a * b = b
(vi) On N, defined by a * b = \(\frac{a+b}{2}\)
Solution:
(i) * on Z⁺ defined by a * b = a – b
Clearly 3 * 6 = 3 – 6 = – 3 ∉ Z⁺ for 3, 6 ∈ Z⁺
Thus a * b ∉ Q ∀ a, b ∈ Z⁺
∴ * is not a binary operation on Z⁺.

(ii) * on Z⁺ defined by a * b = ab ∀ a, b ∈ Z⁺
since product of two positive integers is also a positive integer.
∴ a * b = ab ∈ Z⁺ ∀ a, b ∈ Z⁺
Thus * be a binary operation on Z⁺ .

(iii) Given on Z⁺ define * by a * b = |a – b|
Since the absolute value of difference of two natural numbers is a natural number.
∀ a, b ∈ Z⁺ , a * b = |a – b| ∈ Z⁺
[Since 2 * 3 = | 2 – 3 | = |- 1| = 1 ∈ Z⁺ ]
Similarly for other natural numbers the given property holds.
Thus * be a binary operation on Z⁺ .

(iv) Given operation * on Z⁺ defined by a* b = a
i.e. ∀ a, b ∈ Z⁺ , a * b = a ∈ Z⁺
[since closure property hoids]
∴ * be a binary operation on Z⁺ .

(v) Given operation * on Z⁺ defined by a * b = b
∀ a, b ∈ Z⁺ , a * b = b ∈ Z⁺
[since closure property holds]
∴ * be a binary operation on Z⁺

(vi) Given operation * on N defined by
a * b = \(\frac{a+b}{2}\)
∀ a, b ∈ N, a * b = \(\frac{a+b}{2}\) ∉ N
since 1, 2 ∈ N but \(\frac{1+2}{2}=\frac{3}{2}\) ∉ N
∴ given operation * on N is not binary operation on N.

Question 6 (old).
Let A = {1, 2, 3}, set the composition table for the infimum operation ‘∧’ on A.
Solution:
Given A = {1, 2, 3}
Let ∧ be the binary operation on A defined by a ∧ b = inf {a, b} ∈ a, b ∈ A
The required composition table is given as under:
[∵ Inf. {a, b} = min {a, b}]

Question 7.
Let * on I be a binary operation defined by a * b = 2a + b – 3. Find 3 * 4.
Solution:
Given * be a binary operation on I defined as a*b = 2a + b – 3
∴ 3 * 4 = 2 × 3 + 4 – 3
= 6 + 4- 3 = 7

Question 7 (old).
Let A = {1, 2, 3, 4}, set the composition table for the supermum operation ‘∨’ on A.
Solution:
Given A = {1, 2, 3, 4}
Let ∨ be the binary operation on A defined by
a ∨ b = Sup {a, ft}
The required composition table is given as under :
[Sup {a, b} = a ∨ b = max {a, b}]

Question 8.
Let * on I be a binary operation defined by a * b = 3a + 4b – 2. Find 4 * 5.
Solution:
Given binary operation * on I defined by
a * b = 3a + 4b – 2
Thus, 4 * 5 = 3 × 4 + 4 × 5 – 2
= 12 + 20 – 2
= 32 – 2 = 30

Question 8 (old).
If A = W × W and ‘*’ on A be defined by (a, b) * (c, d) = (a + c, b + d), for all (a, b), (c, d) ∈ A, then show that
(i) ‘*’ is a binary operation on A.
(ii) ‘*’ is commutative on A.
(iii) ‘*’ is associative on A.
Also find the identity element, if any, in A.
Solution:
Given A = W × W and operation * defined on A as
(a, b) * (c, d) = (a + c, b + d) ∀ a, b, c, d ∈ W

Commutativity:
∀ (a, b), (c, d) ∈ A, we have
(a, b) * (c, d) = (a + c, b + d)
(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)
Since commutativity holds under addition on W
∴ (a, b) * (c, d) = (c, d) * (a, b) ∀ (a, b), (c, d) ∈ W × W
Thus * is commutative on A = W x W.

Associativity:
∀ (a, b), (c, d), (e, f) ∈ A = W × W ‘
Here, [(a, b) * (c, d)} * (e, f) = (a + c, b + d) * (e, f)
= (a + c + e, b + d + f)
and (a, b) * [(c, d) * (e, f)] = (a, b) * (c + e, d + f)
= (a + c + e, b + d + f)
Clearly [(a, b) * (c, d)] * (e, f)
= (a, b) * [(c d) * (e, f)]
∀ (a, b), (c, d), (e, f) ∈ A = W × W
Thus * is associative on W × W i.e. A.
Let (x, y) be the identity element in W × W = A
Then (a, b) * (x, y) = (a, b) = (x, y) * (a, b) ∀ (a, b) ∈ A
(a, b) * (x, y) = (a, b)
and (x, y) * (a, b) = (a, b) ∀ (a, b) ∈ A
⇒ (a + x, b + y) = (a, b)
and (x + a, y + b) = (a, b) ∀ (a, b) ∈ A
⇒ a + x = a, b + y = b
and x + a = a, y + b = b ∀ (a, b) ∈ A
⇒ x = 0, y = 0 and x = 0, y = 0
⇒ (x, y) = (0, 0)
Thus (0, 0) be the required identity element in A.

Question 9.
If the binary operation on I is defined by a * b = 2a – 3b, then find
(i) 3 * 7
(ii) 7 * 3
(iii) (2 * 5) * 4.
Solution:
Given binary operation * on I defined by a * b = 2a – 3b
(i) 3 * 7 = 2 × 3 – 3 × 7
= 6 – 21 = – 15

(ii) 7 * 3 = 2 × 7 – 3 × 3
= 14 – 9 = 5

(iii) (2 * 5) * 4 = (2 × 2 – 3 × 5) * 4
= (- 11) * 4
= 2 × (- 11) – 3 × 4
= – 22 – 12 = – 34.

Question 10.
Let a binary operation on R be defined by a * b = \(\sqrt{a^2+b^2}\). Find
(i) 3 * 4
(ii) 3 * (- 4)
Solution:
Given binary operation on R defined by a * b = \(\sqrt{a^2+b^2}\)
(i) 3 * 4 = \(\sqrt{3^2+4^2}\)
= \(\sqrt{25}\) = 5

(ii) 3 * (- 4) = \(\sqrt{3^2+(-4)^2}\)
= \(\sqrt{25}\) = 5

Question 11.
If the binary operation ‘*’ on N is defined by a * b = a^b, then find
(i) 2 * 3
(ii) 3 * 2
(iii) (1 * 2) * 3
(iv) 1 * (2 * 3)
Solution:
Given binary operation * on N defined by a* b = a^b
(i) 2 * 3 = 2³ = 8

(ii) 3 * 2 = 3² = 9

(iii) (1 * 2) * 3 = 1² * 3
= 1 * 3 = 1³ = 1

(iv) 1 * (2 * 3) = 1 * (2³)
= 1 * 8 = 1⁸ = 1

Question 12.
(i) Let * be a binary operation on N given by a * b = HCF (a, b), a, b ∈ N. Write the value of 22 * 4.
Solution:
Given binary operation * on N defined by
a * b = HCF (a, b) ∀ a, b ∈ N
∴ 22 * 4 = HCF (22, 4) = 2

(ii) Let * be a binary operation on N given by a * b = LCM (a, b) for all a, b ∈ N. Find 5 * 7.
Solution:
Given binary operation * on N defined by
a * b = LCM (a, b) ∀ a, b ∈ N
∴ 5 * 7 = LCM (5, 7) = 5 × 7 = 35.

Question 13.
The binary operation ‘*’ R × R → R is defined as a * b = 2a + b. Find (2 * 3) * 4.
Solution:
Given binary operation * on R × R defined by a * b = 2a + b
∴ (2 * 3) * 4 = (2 × 2 + 3) * 4
= 7 * 4
= 2 × 7 + 4
= 18.

Question 13 (old).
If the binary operation * on the set of integers Z, is defined by a * b = a + 3b², then find the value of 2 * 4.
Solution:
Given binary operation * on set of integers Z defined by
a * b = a + 3b²
∴ 2 * 4 = 2 + 3 × 4²
= 2 + 3 × 16
= 2 + 48 = 50

Question 14.
If a * b denotes the larger of a and b and if aob = (a * b) + 3, then find the value of 5 o 10, where * and o are binary operations.
Solution:
Given a * b = (a * b) + 3
where a * b = larger of a and b
∴ 5 * 10 = (5 * 10) + 3
= larger of {5, 10} + 3
= 10 + 3 = 13

Question 15.
Let * be a binary operation, on the set of all non-zero real numbers, given by a * b for all a, b ∈ R – {0}. Find the value of x, given that 2 * (x * 5) = 10.
Solution:
Given binary operation * on R – {0} defined by
a * b = \(\frac{ab}{5}\) ∀ a, b ∈ R – {0}
Given, 2 * \(\left(\frac{x \times 5}{5}\right)\) = 10
⇒ 2 * x = 10
⇒ \(\frac{2 * x}{5}\) = 10
⇒ 2x = 50
⇒ x = 25.

Question 16.
If the binary operation * on the set of integers Z is defined by a * b = a + b – 5, then write the identity clement of * in Z.
Solution:
Given binary operation * on Z defined by
a * b = a + b – 5 ∀ a, b ∈ Z
Let e be the identity element in Z with respect to binary operation *.
Thus, e * a = a = a * e∀ a, b ∈ Z
⇒ e + a – 5 = a
and a = a + e – 5
⇒ e – 5 = 0
⇒ e = 5
Thus 5 be the identity element in Z under *.

Question 17.
If A = {a, b}, then find the number of binary operations on A.
Solution:
Here S = {a, b}
∴ O(S) = 2 = n
Thus required number of binary operations on set S = nⁿ²
= 2²²
= 2⁴ = 16.

Question 18.
If A = {1, 2, 3}, then find the number of binary operations on A.
Solution:
Given S = {a, b, c}
∴ O(S) = 3
We know that, the total number of binary operations on S = nⁿ²
∴ Total number of binary operation on S Containing 3 elements = 3³² = 3⁹

Question 20 (old).
State whether the following statements are true or false. Justify.
(i) For an arbitrary binary * on N, a * a = a, for all a ∈ N.
(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a.
(iii) Every binary operation defined on a set having exactly one element a is necessarily commutative and associative with a as the identity element and a being the inverse of a.
Solution:
(i) For a binary operation denoted by + on N
2 + 2 =4 * 2
∴ given statement is false.

(ii) True, a * (b * c) = (b * c) * a
[∵ * be commutative on N
∴ a * b = b * a ∀ a, b ∈ N]
= (c * b) * a
[∵ b * c = c * b ∀ b, c ∈ N]

(iii) True, Let A = {a} and binary operation * defined on A
a * a = a * a,
a * (a * a) = (a * a) * a ∀ a ∀ A
a * a = a
a behaves as the identity element.
Also a behaves as the inverse of a ≠ 0.

Question 21 (old).
Let be defined on the et {1, 2, 3, 4, 5} by a * b = LCM of a and b. Is * a binary operation ? Justify your answer. (NCERT)
Solution:
Given operation * be defined on A by
a* b = LCM {a, b} where A = (1, 2, 3, 4, 5}
Clearly 3, 5 ∈ A,
3 * 5 = LCM of 3 and 5= 15 ∉ A
∴ given operation * be not a binary operation.

Question 24 (old).
Let A = Q × Q, where Q is the set of rational numbers, and be a binary operations on A defined by (a, b) * (c, d) = (ac, b + ad) for all (a, b), (c, d) ∈ A. Show that the operation * on A is non- commutative but associative.
Solution:
Given * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad)
∀ (a, b), (c, d) ∈ A = Q × Q

Commutativity : ∀ (a, b), (c, d) ∈ A
(a, b) * (c, d) = (ac, b + ad)
(c, d) * (a, b) = (ca, d + cb)
Clearly b + ad ≠ d + cb
∴ (a, b) * (c, d) ≠ (c, d) * (a, b)
Thus * is not commutative on A = Q * Q
Further (1, 2), (3, 4) ∈ A = Q × Q
Here, (1, 2) * (3, 4) = (1 × 3, 2 + 1 × 4) = (3, 6)
(3, 4) * (1, 2) = (3 × 1, 4 + 3 × 2) = (3, 10)
Clearly (1, 2) * (3, 4) ≠ (3, 4) * (1, 2)

Associativity :
∀ (a, b), (c, d), (e, f) ∈ A
[(a, b) * (c, d)} * (e,f) = (ac, b + ad) * (e, f)
= (ace, b + ad + acf)
(a, b) * [(c, d) * (e, f)]
= (a, b) * (ce, d + cf)
= (ace, b + a (d + cf))
= (ace, b + ad + acf)
Thus, [(a, b) * (c, d)] * (e, f) = (a, b) * [(c, d) * (e, f)]
Thus * is associative on A.
Hence, * is not commutative but associative on A.